Now, we consider some specific methods to solve differential equations. The aim is mostly to consider $y$ as a variable dependent on $x$ and solve the equation for $y$. It is, however, common to leave the expressions in implicit forms. One of the most elementary types of differential equations is separable equations.
A first-order differential equation is said to be separable if it can be written in a form where all terms involving the dependent variable $y$ and its differential $dy$ appear on one side of the equation, while all terms involving the independent variable $x$ and its differential $dx$ appear on the other side.
It looks like the following:
$$ \frac{dy}{dx}=\frac{f(x)}{g(y)} $$
or, in a better way,
$$ g(y)dy=f(x)dx $$
Here, we construct our equation in such a form so that we can integrate easily. It is trivial to apply:
$$ \int g(y)dy=\int f(x)dx $$
Letting, by Fundamental Theorem of Calculus, $F$ be an anti-derivative of $f$, and $G$ be an anti-derivative of $g$, we obtain:
$$ G(y)+C_{1}=F(x)+C_{2} $$
Of course, $C_{1}$ and $C_{2}$ are arbitrary constants. Therefore, we can simply collect the constants in one side and say:
$$ G(y)=F(x)+C. $$
Example (1) Solve the following differential equation for $y$ explicitly.
$$ \frac{dy}{dx}=3y^{2}x^{2} $$
Solution: We wish to, firstly, collect the terms involving $dy$ and $y$, and, $dx$ and $x$ in different sides. Here, we obtain
$$ \frac{dy}{3y^{2}}=x^{2}dx. $$
The rest is trivial integration.
$$ \begin{aligned} \int\frac{1}{3y^{2}}dy &= \int x^{2}dx \\ -\frac{1}{3y}+C_{1} &= \frac{x^{3}}{3}+C_{2} \\ -\frac{1}{3y} &= \frac{x^{3}}{3}+C \\ y=y(x) &= -\frac{1}{x^{3}+\tilde{C}} \end{aligned} $$
Example (2) Solve the following differential equation for $y$ explicitly
$$ y^{\prime}=7(x+6)(y^{2}+1) $$
Solution: Now, observe that, in the equation in the example, there exists no explicit differentials. We, however, know that $y^{\prime}$ can be also expressed as $\frac{dy}{dx}$. Here, we note that, as a matter of fact, different notations of derivatives have historical meanings, but, in the course of differential equations, it makes it much easier to use Leibniz notation, that is, $\frac{dy}{dx}$. If the reader wants further information about the topic, $y^{\prime}$ is referred as Lagrange’s notation, and $\dot{y}$ is referred as Newton’s notation (of derivatives).
Now, the motivation to solve the equation stays the same. We have:
$$ \frac{dy}{dx}=7(x+6)(y^{2}+1) $$
which is followed by
$$ \frac{dy}{y^{2}+1}=7(x+6)dx $$
Then, we may apply integration as follows,
$$ \begin{aligned} \int\frac{1}{y^{2}+1}dy &= \int7(x+6)dx \\ \arctan(y) &= \frac{7x^{2}}{2}+42x+C \end{aligned} $$
Here, we note that it is natural to leave $\arctan(y)$, since the transformation from $\arctan(y)$ to $y$ requires a careful analysis of the interval of $y,$ as the inverse trigonometric functions are defined on restricted intervals.